An 18.0-kg box is released on the 37.0 degree incline andaccelerations down the

Question

An 18.0-kg box is released on the 37.0 degree incline andaccelerations down the incline at 0.27 m/s^2. Find thefiriction force impeding its motion. How large is the coefficientof kinetic friction. When I get down to Uk=1/cos37.0 degrees[sin 37degree - 0.27/9.80] Could you show me how to work that on mycalculator and/or how to work the problem out period? I come upwith some off the wall number when I key it in so I'm doingsomething wrong. Thank you for your help.
An 18.0-kg box is released on the 37.0 degree incline andaccelerations down the incline at 0.27 m/s^2. Find thefiriction force impeding its motion. How large is the coefficientof kinetic friction. When I get down to Uk=1/cos37.0 degrees[sin 37degree - 0.27/9.80] Could you show me how to work that on mycalculator and/or how to work the problem out period? I come upwith some off the wall number when I key it in so I'm doingsomething wrong. Thank you for your help.

Explanation / Answer

Uk=1/cos37.0 degrees[sin 37 degree - 0.27/9.80]    0.27/9.8 = 0.027 sin37 = 0.60 1/cos37 = 1/0.79 = 1.25               k = 1.25 ( 0.60 - 0.027) = 0.71 Fk = mg cos37 plug in values Fk = ------

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